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JEE Main 202523 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
23 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Thermodynamics (C)

JEE Main 2025 · 23 January · Morning Shift · Chemistry · Thermodynamics (C)

^1 The standard enthalpy and standard entropy of decomposition of...

1{ }^1 The standard enthalpy and standard entropy of decomposition of N2O4\mathrm{N}_2 \mathrm{O}_4 to NO2\mathrm{NO}_2 are 55.0 kJ mol−155.0 \mathrm{~kJ} \mathrm{~mol}^{-1} and 175.0 J/K/mol175.0 \mathrm{~J} / \mathrm{K} / \mathrm{mol} respectively. The standard free energy change for this reaction at 25∘C25^{\circ} \mathrm{C} in J mol−1\mathrm{mol}^{-1} is ______ (Nearest integer)

Correct answer

0

Step-by-step solution

ΔHrxno=55 kJ/mol,T=298 KΔ Srxno=175 J/molΔGrxno=ΔHrxno−TΔSrxn∘⇒ΔGrxno=55000 J/mol−298×175 J/mol⇒ΔGrxno=55000−52150⇒ΔGrxno=2850 J/mol\begin{aligned} & \Delta \mathrm{H}_{\mathrm{rxn}}^{\mathrm{o}}=55 \mathrm{~kJ} / \mathrm{mol}, \quad \mathrm{T}=298 \mathrm{~K} \\ & \Delta \mathrm{~S}_{\mathrm{rxn}}^{\mathrm{o}}=175 \mathrm{~J} / \mathrm{mol} \\ & \Delta \mathrm{G}_{\mathrm{rxn}}^{\mathrm{o}}=\Delta \mathrm{H}_{\mathrm{rxn}}^{\mathrm{o}}-\mathrm{T} \Delta \mathrm{S}_{\mathrm{rxn}}^{\circ} \\ & \Rightarrow \Delta \mathrm{G}_{\mathrm{rxn}}^{\mathrm{o}}=55000 \mathrm{~J} / \mathrm{mol}-298 \times 175 \mathrm{~J} / \mathrm{mol} \\ & \Rightarrow \Delta \mathrm{G}_{\mathrm{rxn}}^{\mathrm{o}}=55000-52150 \\ & \Rightarrow \Delta \mathrm{G}_{\mathrm{rxn}}^{\mathrm{o}}=2850 \mathrm{~J} / \mathrm{mol}\end{aligned}
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